Force bc it says the ability to make stuff happen
Answer: 0.067 s
Explanation:s = Ut + 1/2at^2
0.6 = 9t + 0.5 *10 *t^2
Where a = g =10m/s/s
Solving the quadratic equation
5t^2 + 9t - 0.6=0,
t= 0.067 s and - 1.7 s
Of which 0.067 s is a valid time
Answer:
(a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Explanation:
Given that,
Power factor = 0.6
Power = 600 kVA
(a). We need to calculate the reactive power
Using formula of reactive power
...(I)
We need to calculate the
Using formula of
Put the value into the formula
Put the value of Φ in equation (I)
(b). We draw the power triangle
(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95
Using formula of reactive power
We need to calculate the difference between Q and Q'
Put the value into the formula
Hence, (a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Answer:
a=F/m
a=12N/3kg (here newton can be written as kgm/s^2 so kg will be cancelled)
a=4m/s^2
Explanation:
Your diagram should include four forces:
• the box's weight, pointing down (magnitude <em>w</em> = 43.2 N)
• the normal force, pointing up (mag. <em>n</em>)
• the applied force, pointing the direction in which the box is sliding (mag. <em>p</em> = 6.30 N, with <em>p</em> for "pull")
• the frictional force, pointing oppoiste the applied force (mag. <em>f</em> )
The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have
<em>n</em> + (-<em>w</em>) = 0
and
<em>p</em> + (-<em>f</em> ) = 0
So then the forces have magnitudes
<em>w</em> = 43.2 N
<em>n</em> = <em>w</em> = 43.2 N
<em>p</em> = 6.30 N
<em>f</em> = <em>p</em> = 6.30 N